Question 12 Marks
Verify : $x^3 + y^3 = (x + y)(x^2 - xy + y^2)$
AnswerWe know that
$(x + y)^3 = x^3 + y^3 + 3xy(x + y)$ Using Identity $(a + b)^3= a^3+ b^3+ 3ab (a + b)$
$\Rightarrow x^3 + y^3 = (x + y)^3 - 3xy(x + y)$
$\Rightarrow x^3 + y^3 = (x + y){(x + y)^2 - 3xy}$
$\Rightarrow x^3 + y^3 = (x + y)(x^2 + 2xy + y^2 - 3xy)$ Using Identity $(a + b)^2= a^2+ 2ab + b^2$
$\Rightarrow x^3 + y^3= (x + y)(x^2 - xy + y^2)$
View full question & answer→Question 22 Marks
Verify : $x^3- y^3 = (x - y)(x^2 + xy + y^2)$
AnswerWe know that
$(x - y)^3 = x^3 - y^3 - 3xy(x - y)$ Using Identity $(a – b)^3= a^3– b^3– 3ab (a – b)$
$\Rightarrow x^3 - y^3 = (x - y)^3 + 3xy(x - y)$
$\Rightarrow x^3 - y^3 = (x - y){(x - y)^2 + 3xy}$
$\Rightarrow x^3 - y^3 = (x - y)(x^2 - 2xy + y^2 + 3xy)$
$\Rightarrow x^3 - y^3 = (x - y)(x^2 + xy+ y^2)$
View full question & answer→Question 32 Marks
Factorise : $27p^3 - \frac{1}{216} - \frac{9}{2} p^2 + \frac{1}{4}p.$
Answer$27p^3 - \frac{1}{216} - \frac{9}{2}p^2 + \frac{1}{4}p$
$= (3 p)^{3}-\left(\frac{1}{6}\right)^{3}-3(3 p)\left(\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
$= \left(3 p-\frac{1}{6}\right)^{3}$
(Using Identity) $(a – b)^3$
$= a^3– b^3– 3ab (a – b)$
$= \left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)\left(3 p-\frac{1}{6}\right)$
View full question & answer→Question 42 Marks
Factorise : $8a^3 + b^3+ 12a^2b + 6ab^2$
Answer$8a^3 + b^3 + 12a^2b + 6ab^2$
$= (2a)^3 + (b)^3 + 3(2a)(b)(2a + b)$
$= (2a+ b)^3$
(Using Identity) $(a + b)^3$
$= a^3+ b^3+ 3ab (a + b)$
$= (2a + b)(2a + b)(2a + b)$
View full question & answer→Question 52 Marks
Evaluate the using suitable identity: $(998)^3$
Answer$(998)^3$
$= (1000 - 2)^3$
$= (1000)^3 - (2)^2 - 3(1000)(2)(1000 - 2)$
(Using Identity) $(a – b)^3$
$= a^3 – b^3 – 3ab (a – b)$
$= 1000000000 - 8 - 6000(1000 - 2)$
$= 1000000000 - 8 - 6000000 + 12000$
$= 994011992$
View full question & answer→Question 62 Marks
Using suitable identities Evaluate : $(102)^3$
Answer$(102)^3$
$= (100 + 2)^3$
$= (100)^3 + (2)^3 + 3(100)(2)(100 + 2)$
(Using Identity) $(a + b)^3 = a^3 + b^3 + 3ab (a + b)$
$= 1000000 + 8 + 600(100 + 2)$
$= 1000000 + 8 + 60000 + 1200$
$= 1061208$
View full question & answer→Question 72 Marks
Write the cube in expanded form : $\left(x-\frac{2}{3} y\right)^{3}$
Answer$\left(x-\frac{2}{3} y\right)^{3}$
$= x^{3}-\left(\frac{2}{3} y\right)^{3}-3(x)\left(\frac{2}{3} y\right)\left(x-\frac{2}{3} y\right)$
(Using Identity) $(a – b)^3$
$= a^3 – b^3 – 3ab (a – b)$
$= x^{3}-\frac{8}{27} y^{3}-2 x y\left(x-\frac{2}{3} y\right)$
$= x^{3}-\frac{8}{27} y^{3}-2 x^{2} y+\frac{4}{3} x y^{2}$
$= x^3 - 2x^2y + \frac{4}{3}xy^2 - \frac{8}{27}y^3$
View full question & answer→Question 82 Marks
Factorize:
$2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$
Answer$2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$
We need to factorize the expression $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$
The expression $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz$ can also be written as
${\left( { - \sqrt 2 x} \right)^2} + {\left( y \right)^2} + {\left( {2\sqrt 2 z} \right)^2} + 2 \times \left( { - \sqrt 2 x} \right) \times y + 2 \times y \times \left( {2\sqrt 2 z} \right) + 2 \times \left( {2\sqrt 2 z} \right) \times \left( { - \sqrt 2 x} \right).$
We can observe that, we can apply the identity ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca$ with respect to the expression
${\left( { - \sqrt 2 x} \right)^2} + {\left( y \right)^2} + {\left( {2\sqrt 2 z} \right)^2} + 2 \times \left( { - \sqrt 2 x} \right) \times y + 2 \times y \times \left( {2\sqrt 2 z} \right) + 2 \times \left( {2\sqrt 2 z} \right) \times \left( { - \sqrt 2 x} \right)$,to get
${\left( { - \sqrt 2 x + y + 2\sqrt 2 z} \right)^2}$
Therefore, we conclude that after factorizing the expression $2{x^2} + {y^2} + 8{z^2} - 2\sqrt 2 xy + 4\sqrt 2 yz - 8xz $ we get ${\left( { - \sqrt 2 x + y + 2\sqrt 2 z} \right)^2}$.
View full question & answer→Question 92 Marks
Evaluate the product without multiplying directly: $104 \times 96$
Answer$104 \times 96$
$104 \times 96{\text{ can also be written as}}\left( {100 + 4} \right)\left( {100 - 4} \right).$ We can observe that, we can apply the identity $\left( {x + y} \right)\left( {x - y} \right) = {x^2} - {y^2}$ with respect to the expression $\left( {100 + 4} \right)\left( {100 - 4} \right)$,to get
$\left( {100 + 4} \right)\left( {100 - 4} \right) = {\left( {100} \right)^2} - {\left( 4 \right)^2}$ $= 10000 - 16$
= 9984
Therefore, we conclude that the value of the product $104 \times 96 $ is 9984
View full question & answer→Question 102 Marks
Evaluate the product without multiplying directly: $95 \times 96$
Answer$95 \times 96$
$95 \times 96{\text{ can also be written as}}\left( {100 - 5} \right)\left( {100 - 4} \right)$ We can observe that we can apply the identity $\left( {x + a} \right)\left( {x + b} \right) = {x^2} + \left( {a + b} \right)x + ab$ Here a = -5 and b = -4
$\left( {100 - 5} \right)\left( {100 - 4} \right) = {\left( {100} \right)^2} + \left[ {\left( { - 5} \right) + \left( { - 4} \right)} \right]\left( {100} \right) + \left( { - 5} \right) \times \left( { - 4} \right)$
$= 10000 - 900 + 20$
= 9120 Therefore, we conclude that the value of the product $95 \times 96$ is 9120
View full question & answer→Question 112 Marks
What is the possible expression for the dimension of the cuboid whose volume is $12k{y^2} + 8ky - 20k$
Answer${\text{Volume : 12}}k{y^2} + 8ky - 20k$
The expression ${\text{12}}k{y^2} + 8ky - 20k $ can also be written as $ k\left( {{\text{12}}{y^2} + 8y - 20} \right).$
$k\left( {{\text{12}}{y^2} + 8y - 20} \right) = k\left( {{\text{12}}{y^2} - 12y + 20y - 20} \right)$ $\, = k\left[ {12y\left( {y - 1} \right) + 20\left( {y - 1} \right)} \right]$
$ = k\left( {12y + 20} \right)\left( {y - 1} \right)$
$= 4k \times \left( {3y + 5} \right) \times \left( {y - 1} \right).$
Therefore, we can conclude that a possible expression for the dimension of a cuboid of volume ${\text{12}}k{y^2} + 8ky - 20k \ is \ 4k\ ,\left( {3y + 5} \right){\text{ and }}\left( {y - 1} \right)$
View full question & answer→Question 122 Marks
Give possible expression for the length and breadth of the rectangle, in which the area is $35{y^2} + 13y - 12$
Answer${\text{Area : }}35{y^2} + 13y - 12$
The expression $35{y^2} + 13y - 12$ can also be written as, $35y^2 + 28y - 15y - 12$
$= 35{y^2} + 28y - 15y - 12$
$= 7y\left( {5y + 4} \right) - 3\left( {5y + 4} \right)$
$= \left( {7y - 3} \right)\left( {5y + 4} \right)$
Therefore, we can conclude that a possible expression for the length and breadth of a rectangle of area $35{y^2} + 13y - 12$ is Length $= 7y - 3$ and Breadth $= 5y + 4$
View full question & answer→Question 132 Marks
Give possible expression for the length and breadth of the rectangle, in which the area is $25{a^2} - 35a + 12$
Answer${\text{Area : }}25{a^2} - 35a + 12$
The expression $25a^2 - 35a + 12$ can also written as $25a^2 - 15a - 20a + 12$
$25{a^2} - 15a - 20a + 12$
$= 5a\left( {5a - 3} \right) - 4\left( {5a - 3} \right)$
$\, = \left( {5a - 4} \right)\left( {5a - 3} \right).$
Therefore, we can conclude that a possible expression for the length and breadth of a rectangle of area $25{a^2} - 35a + 12$ is ${\text{Length}}$
$= \left( {5a - 4} \right){\text{ and Breadth}}$
$= \left( {5a - 3} \right)$
View full question & answer→Question 142 Marks
Use suitable identity to find the product: $(3 - 2x)(3 + 2x)$
Answer$(3 - 2x)(3 + 2x) = (3)^2 - (2x)^2$
Using Identity $a^2- b^2$
$= (a- b)(a+ b) $
$= 9 - 4x^2$
View full question & answer→Question 152 Marks
Without actually calculating the cube, find the value of $(-12)^3 + (7)^3 + (5)^3.$
Answer$a^3 + b^3 + c^3 = 3abc$
If $a + b + c = 0$
Given: $a = -12, b = 7, c = 5$
And $a + b + c = -12 + 7 + 5 = 0$
Then,
$(-12)^3 + (7)^3 + (5)^3$
$= 3 \times - 12 \times 7 \times 5$
$= -1260$
View full question & answer→Question 162 Marks
Use suitable identity to find the product: $(x + 8)(x - 10)$
Answer$(x + 8)(x - 10)$
$= (x + 8){x + (-10)}$
(Using Identity) $(x + a)(x + b)$
$= x^2 + (a + b) x + ab$
$= x^2 + {8 + (-10)}x + (8)(-10)$
$= x^2 - 2x - 80$
View full question & answer→Question 172 Marks
Factorise :$27y^3 + 125z^3$
Answer$27y^3 + 125z^3$
$= (3y)^3 + (5z)^3 = (3y + 5z){(3y)^2 - (3y)(5z) + (5z)^2}$
$= (3y + 5z)(9y^2 - 15yz + 25z^2)$
View full question & answer→Question 182 Marks
Find the remainder when $x^3 - ax^2 + 6x - a$ is divided by $x - a.$
AnswerLet $p(x) = x^3 - ax^2+ 6x - a$
$x - a = 0$
$\Rightarrow x = a$
$\therefore$ Remainder $= (a)^3 - a(a)^2 + 6(a) - a$
$= a^3 - a^3 + 6a - a$
$= 5a$
View full question & answer→Question 192 Marks
Find the remainder when $ {x^3} + 3{x^2} + 3x + 1$ is divided by $x - \frac{1}{2}$
Answer$x - \frac{1}{2}$
We need to find the zero of the polynomial $x - \frac{1}{2}$
$\begin{gathered} x - \frac{1}{2} = 0{\text{ }} \hfill \\ \Rightarrow {\text{ }}x = \frac{1}{2} \hfill \\ \end{gathered} $
While applying the remainder theorem, we need to put the zero of the polynomial $x - \frac{1}{2}$ in the polynomial ${x^3} + 3{x^2} + 3x + 1$, to get
$p\left( x \right) = {x^3} + 3{x^2} + 3x + 1$
$p\left( {\frac{1}{2}} \right) = {\left( {\frac{1}{2}} \right)^3} + 3{\left( {\frac{1}{2}} \right)^2} + 3\left( {\frac{1}{2}} \right) + 1$
$ = \frac{1}{8} + 3\left( {\frac{1}{4}} \right) + \frac{3}{2} + 1$ = ${1 \over 8} + {3 \over 4} + {3 \over 2} + 1$
$= \frac{{1 + 6 + 12 + 8}}{8}$
$\, = \frac{{27}}{8}$
Therefore, we conclude that on dividing the polynomial ${x^3} + 3{x^2} + 3x + 1 by \ x - \frac{1}{2}$ we will get the remainder as $\frac{{27}}{8}$
View full question & answer→Question 202 Marks
Verify $x = - \frac{1}{2}$ are zeroes of the polynomial $p\left( x \right) = 2x + 1$
Answer$p\left( x \right) = 2x + 1,\, x = - \frac{1}{2}$
We need to check whether $p\left( x \right) = 2x + 1,\, x = - \frac{1}{2}$ is equal to zero or not, i.e., $p\left( {{-1 \over 2}} \right)$ is equal to 0 or not.
$p\left( { - \frac{1}{2}} \right) = 2\left( { - \frac{1}{2}} \right) + 1\,\, = - 1 + 1\,\, = 0$ Since $p\left( {{-1 \over 2}} \right)$ = 0
Therefore, $x = - \frac{1}{2}$is a zero of the polynomial $p\left( x \right) = 2x + 1$
View full question & answer→Question 212 Marks
Verify $x = - \frac{m}{l}$ are zeroes of the polynomial $p\left( x \right) = lx + m$
Answer$p\left( x \right) = lx + m,\, x = - \frac{m}{l}$ We need to check whether $p\left( x \right) = lx + m{\text{ at }}x = - \frac{m}{l}$ is equal to zero or not, i.e., $p\left( {{{ - m} \over l}} \right)$ is equal to zero or not.
$p\left( { - \frac{m}{l}} \right) = l\left( { - \frac{m}{l}} \right) + m\,\, = -m + m\,\, = 0$
Therefore, $x = - \frac{m}{l}$ is a zero of the polynomial $p\left( x \right) = lx + m$
View full question & answer→Question 222 Marks
Verify $x = 0$ are zeroes of the polynomial $p\left( x \right) = {x^2}$
Answer$p\left( x \right) = {x^2},\,\,\,\,x = 0$
We need to check whether $p\left( x \right) = {x^2}{\text{ at }}x = 0$ is equal to zero or not, i.e., $p\left( 0 \right)$ is equal to zero or not.
$p\left( 0 \right) = {\left( 0 \right)^2}\, = 0$
Therefore, we can conclude that $x = 0$ is a zero of the polynomial $p\left( x \right) = {x^2}$
View full question & answer→Question 232 Marks
Verify $x = - 1,2$ are zeroes of the polynomial $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right)$
Answer$p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right),\, x = - 1,2$ We need to check whether $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right){\text{ at }}x = - 1,2$ is equal to zero or not, i.e., $p\left( { - 1} \right)$ and $p\left( 2 \right)$ is equal to zero or not.
At $x = - 1$, $p\left( { - 1} \right) = \left( { - 1 + 1} \right)\left( { - 1 - 2} \right)\,\, = \left( 0 \right)\left( { - 3} \right)\,\, = 0$
At $x = 2$, $p\left( 2 \right) = \left( {2 + 1} \right)\left( {2 - 2} \right)\,\, = \left( 3 \right)\left( 0 \right)\,\, = 0$
Therefore, $x = - 1,2$ are the zeros of the polynomial $p\left( x \right) = \left( {x + 1} \right)\left( {x - 2} \right)$
View full question & answer→Question 242 Marks
Verify $x = - 1,1$ are zeroes of the polynomial $p\left( x \right) = {x^2} - 1$
Answer$p\left( x \right) = {x^2} - 1,\, x = - 1,1$
We need to check whether $p\left( x \right) = {x^2} - 1{\text{ at }}x = - 1,1$ is equal to zero or not, i.e., $p\left( { - 1} \right)$ and $p\left( { 1} \right)$ is equal to zero or not.
At $x = - 1$
$p\left( { - 1} \right) = {\left( { - 1} \right)^2} - 1\,\, = 1 - 1\,\, = 0$
At $x = 1$
$p\left( 1 \right) = {\left( 1 \right)^2} - 1\,\, = 1 - 1\,\, = 0$
Therefore ,$x = - 1,1$ are the zeros of the polynomial $p\left( x \right) = {x^2} - 1$
View full question & answer→Question 252 Marks
Verify $x = \frac{4}{5}$ are zeroes of the polynomial $p\left( x \right) = 5x - \pi$
Answer$p\left( x \right) = 5x - \pi ,\, x = \frac{4}{5}$
We need to check whether $p\left( x \right) = 5x - \pi {\text{ at }}x = \frac{4}{5}$ is equal to zero or not, i.e., $p\left( {{4 \over 5}} \right)$ is equal to zero or not.
$p\left( {\frac{4}{5}} \right) = 5\left( {\frac{4}{5}} \right) - \pi \,\, = 4 - \pi$
Therefore, $x = \frac{4}{5}$ is not a zero of the polynomial $ p\left( x \right) = 5x - \pi $
View full question & answer→Question 262 Marks
Verify $x = - \frac{1}{3}$ are zeroes of the polynomial $p\left( x \right) = 3x + 1$
Answer$p\left( x \right) = 3x + 1, x = - \frac{1}{3}$
We need to check whether $p\left( x \right) = 3x + 1{\text{ at }}x = - \frac{1}{3}$ is equal to zero or not.
$p\left( { - \frac{1}{3}} \right) = 3x + 1 = 3\left( { - \frac{1}{3}} \right) + 1\, = - 1 + 1\, = 0$
Therefore, we can conclude that $x = - \frac{1}{3}$ is a zero of the polynomial $p\left( x \right) = 3x + 1$
View full question & answer→Question 272 Marks
Find p(0), p(1) and p(2) for the polynomial: p(x) = (x – 1)(x + 1)
Answerp(x) = (x – 1)(x + 1)
$\therefore$ p(0) = (0 – 1)(0 + 1) = (–1)(1) = –1
p(1) = (1 – 1)(1 + 1) = (0)(2) = 0,
p(2) = (2 – 1)(2 + 1) = (1)(3) = 3
View full question & answer→Question 282 Marks
Find p(0), p(1) and p(2) of the polynomial: $p ( t ) = 2 + t + 2 t ^ { 2 } - t ^ { 3 }$
AnswerAccording to the question, $p ( t ) = 2 + t + 2 t ^ { 2 } - t ^ { 3 }$
$p ( 0 ) = 2 + ( 0 ) + 2 ( 0 ) ^ { 2 } - ( 0 ) ^ { 3 } = 2$
$p ( 1 ) = 2 + ( 1 ) + 2 ( 1 ) ^ { 2 } - ( 1 ) ^ { 3 } = 2 + 1 + 2 - 1 = 4$
$p ( 2 ) = 2 + ( 2 ) + 2 ( 2 ) ^ { 2 } - ( 2 ) ^ { 3 } = 4 + 8 - 8 = 4$
View full question & answer→Question 292 Marks
Verify whether $2$ and $0$ are zeroes of the polynomial $x^2 – 2x$
AnswerLet $p(x) = x^2 – 2x$
Then $p(2) = 2^2 – 4 = 4 – 4 = 0$
and $p(0) = 0 – 0 = 0$
Hence, $2$ and $0$ are both zeroes of the polynomial $x^2 – 2x.$
View full question & answer→Question 302 Marks
Factorise : $8x^3 + y^3 + 27z^3 – 18xyz$
AnswerHere, we have
$8x^3 + y^3 + 27z^3 - 18xyz$
$= (2x)^3 + y^3 + (3z)^3 – 3(2x)(y)(3z)$
$= (2x + y + 3z)[(2x)^2 + y^2 + (3z)^2 – (2x)(y) – (y)(3z) – (2x)(3z)]$
$= (2x + y + 3z) (4x^2 + y^2 + 9z^2 - 2xy - 3yz - 6xz)$
This is the required factorisation.
View full question & answer→Question 312 Marks
Factorise: $8x^3 + 27y^3 + 36x^2y + 54xy^2$
AnswerThe given expression can be written as
$(2x)^3 + (3y)^3 + 3(4x^2)(3y) + 3(2x)(9y^2)$
$= (2x)^3 + (3y)^3 + 3(2x)^2(3y) + 3(2x)(3y)^2$
$= (2x + 3y)^3$ (Using Identity) $(x + y)^3$
$= x^3 + y^3 + 3xy (x + y)$
$= (2x + 3y)(2x + 3y)(2x + 3y)$
This is the required factorisation.
View full question & answer→Question 322 Marks
Evaluate: $(104)^3$ using a suitable identity.
Answer$(104)^3 = (100 + 4)^3$
Using identity $(x + y)^3 = x^3 + 3xy(x + y) + y^3$
We get,
$(100 + 4)^3 = (100)^3 + 3 \times 100 \times 4 (100 + 4) + 4^3$
$= 10,00 000 + 1,200 \times 104 + 64$
$= 10,00,000 + 1,24,800 + 64$
$= 11,24,864$
View full question & answer→Question 332 Marks
Write $(5p – 3q)^3$ in the expanded form.
AnswerComparing the given expression with $(x – y)^3,$
we find that $x = 5p, y = 3q.$
So, using Identity $(x - y)^{3 }= x^3 – 3x^2y + 3xy^2 – y^3,$
we have: $(5p – 3q)^3 = (5p)^3 – (3q)^3 – 3(5p)(3q)(5p – 3q)$
$= 125p^3 – 27q^3 – 225p^2q + 135pq^2$
This is the required expansion.
View full question & answer→Question 342 Marks
Write $(3a + 4b)^3$ in the expanded form.
AnswerComparing the given expression with $(x + y)^3 ,$
we find that $x = 3a$ and $y = 4b.$
So, using Identity $(x + y)^3 = x^3 + y^3 + 3xy (x + y),$
we have: $(3a + 4b)^3 = (3a)^3 + (4b)^3 + 3(3a)(4b)(3a + 4b)$
$= 27a^3 + 64b^3 + 108a^2b + 144ab^2$
This is the required expansion.
View full question & answer→Question 352 Marks
Find the value of $q(y) = 3y^3 - 4y + \sqrt{11}$ at $y = 2.$
AnswerWe have, $q(y) = 3y^3 - 4y + \sqrt{11}$
On put $y = 2$ in $q(y),$ we get
$q(2) = 3(2)^3 - 4(2)+\sqrt{11}$
$= 3 \times8 - 8 + \sqrt{11}$
$= 24 - 8 + \sqrt{11}$
$= 16 + \sqrt{11}$
View full question & answer→Question 362 Marks
Factorise : $4x^2 + y^2 + z^2 – 4xy – 2yz + 4xz$
AnswerWe have,
$4x^2 + y^2 + z^2 – 4xy – 2yz + 4xz = (2x)^2 + (–y)^2 + (z)^2 + 2(2x)(–y) + 2(–y)(z) + 2(2x)(z)$
$= [2x + (–y) + z]^2$
$= (2x – y + z)^2$
$= (2x – y + z)(2x – y + z)$
This is the required factorisation.
View full question & answer→Question 372 Marks
Factorise : $49a^2 + 70ab + 25b^2$
AnswerHere you can see that $49a^2 = (7a)^2, 25b^2 = (5b)^2, 70ab = 2(7a) (5b)$
Comparing with $x^2 + 2xy + y^2,$ we observe that $x = 7a$ and $y = 5b.$
Using Identity $(x + y)^2$
$= x^2 + 2xy + y,$ we get
$49a^2 + 70ab + 25b^2$
$= (7a + 5b)^2$
$= (7a + 5b) (7a + 5b)$
This is the required factorisation.
View full question & answer→Question 382 Marks
Evaluate $105 \times 106$ without multiplying directly.
AnswerFirstly, write $105$ as $100 + 5$ and $106$ as $100 + 6.$
$\therefore 105 \times 106 = (100 + 5) \times (100 + 6)$
$= (100)^2 + (5 + 6)100 + (5 \times 6)$
$[\because (x + a)(x + b) = x^2 + (a + b)x + ab]$
$= 10000+ 11\times100 + 30$
$= 10000 + 1100 + 30 = 11130$
View full question & answer→Question 392 Marks
Factorise $y^2 – 5y + 6$ by using the Factor Theorem.
AnswerLet $p(y) = y^2 – 5y + 6.$
The factors of $6$ are $1, 2$ and $3.$
Now, $p(2) = 22 – (5 \times 2) + 6 = 0$
So, $y – 2$ is a factor of $p(y).$
Also, $p(3) = 32 – (5 \times 3) + 6 = 0$
So, $y – 3 $ is also a factor of $y^2 – 5y + 6$
Therefore, $y^2 – 5y + 6 = (y – 2)(y – 3)$ This is the required factorisation.
View full question & answer→Question 402 Marks
Factorise $6x^2 + 17x + 5$ by splitting the middle term
AnswerIf we can find two numbers $p$ and $q$ such that $p + q = 17$ and$ pq = 6 \times 5 = 30,$ then we can get the factors
So, let us look for the pairs of factors of $30.$ Some are $1$ and $30, 2$ and $15, 3$ and $10, 5$ and $6.$ Of these pairs$, 2$ and $15$ will give us$ p + q = 17.$
So$, 6x^2 + 17x + 5$
$= 6x^2 + (2 + 15)x + 5$
$= 6x^2 + 2x + 15x + 5$
$= 2x(3x + 1) + 5(3x + 1)$
$= (3x + 1) (2x + 5)$
This is the required factorisation.
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Find the value of $k, if \ x – 1$ is a factor of $4x^3 + 3x^2 – 4x + k.$
AnswerAs $x – 1$ is a factor of $p(x) = 4x^3 + 3x^2 \ – 4x + k, $
$\therefore,p(1) = 0$
Now$, p(1) = 4(1)^3 + 3(1)^2 \ – 4(1) + k$
$So, 4 + 3\ – 4 + k = 0$
$i.e., k = \ –3$
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